Problem: Which of the following numbers is a multiple of 9? ${48,53,68,81,103}$
Answer: The multiples of $9$ are $9$ $18$ $27$ $36$ ..... In general, any number that leaves no remainder when divided by $9$ is considered a multiple of $9$ We can start by dividing each of our answer choices by $9$ $48 \div 9 = 5\text{ R }3$ $53 \div 9 = 5\text{ R }8$ $68 \div 9 = 7\text{ R }5$ $81 \div 9 = 9$ $103 \div 9 = 11\text{ R }4$ The only answer choice that leaves no remainder after the division is $81$ $ 9$ $9$ $81$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $9$ are contained within the prime factors of $81$ $81 = 3\times3\times3\times3 9 = 3\times3$ Therefore the only multiple of $9$ out of our choices is $81$. We can say that $81$ is divisible by $9$.